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Sagot :
Answer:
a = 7.29 m / s², T = 0.40 N
Explanation:
To solve this exercise we must apply Newton's second law to each body
The needle
W -T = m a
mg - T = ma
The spool, which we will approach by a cylinder
Σ τ = I α
T R = I α
the moment of inertia of a cylinder with an axis through its center is
I = ½ M R²
angular and linear variables are related
a = α R
α = a / R
we substitute
T R = (½ M R²) a / R
T = ½ M a
we write our system of equations together
mg - T = m a
T = ½ M a
we solve
m g = (m + ½ M) a
a = [tex]\frac{m}{m + \frac{1}{2} M} \ g[/tex]
let's calculate
a = [tex]\frac{0.160}{0.160 + \frac{1}{2} 0.110} \ 9.8[/tex]
a = 7.29 m / s²
now we can look for the tension
T = ½ M a
T = ½ 0.110 7.29
T = 0.40 N
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